`f:R to R" given by "f(x)=x+sqrt(x^(2))`,is

Options

A. injective
B. surjective
C. bijective
D. none of these

Explanation

To solve the problem, we need to analyze the function \( f: \mathbb{R} \to \mathbb{R} \) given by

\[ f(x) = x + \sqrt{x^2} \]

### Step 1: Simplify the function
The term \( \sqrt{x^2} \) can be simplified. For any real number \( x \), \( \sqrt{x^2} = |x| \) (the absolute value of \( x \)). Therefore, we can rewrite the function as:

\[ f(x) = x + |x| \]

### Step 2: Determine the behavior of \( f(x) \) based on the sign of \( x \)
Next, we will analyze the function based on the value of \( x \):

1. **When \( x \geq 0 \)**:
- Here, \( |x| = x \).
- Thus, \( f(x) = x + x = 2x \).

2. **When \( x < 0 \)**:
- Here, \( |x| = -x \).
- Thus, \( f(x) = x - x = 0 \).

So we can summarize the function as:

\[ f(x) = \begin{cases} 2x & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases} \]

### Step 3: Analyze injectivity (one-to-one)
A function is injective if different inputs produce different outputs.

- For \( x < 0 \), \( f(x) = 0 \) for all \( x < 0 \). This means all negative inputs map to the same output (0), indicating that the function is not injective.
- For \( x \geq 0 \), \( f(x) = 2x \) is a linear function with a positive slope, which is injective in this interval.

Since the function is not injective over the entire domain, it is not a one-to-one function.

### Step 4: Analyze surjectivity (onto)
A function is surjective if every element in the codomain has a pre-image in the domain.

- The range of \( f(x) \):
- For \( x < 0 \), \( f(x) = 0 \).
- For \( x \geq 0 \), \( f(x) = 2x \) which ranges from \( 0 \) to \( +\infty \).

Thus, the range of \( f(x) \) is \( [0, +\infty) \). Since the codomain is \( \mathbb{R} \) (which includes negative numbers), not every real number has a pre-image. Therefore, the function is not surjective.

### Conclusion
Since the function \( f(x) \) is neither injective nor surjective, the correct answer is:

**None of this.**